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Application of Derivatives Test - 44

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Application of Derivatives Test - 44
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  • Question 1
    1 / -0
    If the tangent at $$(x_{1}, y_{1})$$ to the curve $$x^{3}+y^{3}=a^{3}$$ meets the curve again at $$(x_{2}, y_{2})$$ then
    Solution
    Given $$x^3+y^3=a^3$$
    The derivative is
            $$\dfrac{dy}{dx}=-\dfrac{x^2}{y^2}.....(1)$$
    Therefore, slope of tangent at $$(x_1,y_1)$$ is
               $$-\dfrac{x_1^2}{y_1^2}.........(2)$$
    The tangent passes through $$(x_2,y_2)$$, therefore the slope of tangent is also given by 
                $$\dfrac{y_2-y_1}{x_2-x_1}.......(3)$$
    Comparing the two slope equations we get
    $$\dfrac{y_2-y_1}{x_2-x_1}=-\dfrac{x_1^2}{y_1^2}$$

    $$\dfrac{y_2^3-y_1^3}{x_2^3-x_1^3}\times\dfrac{x_1^2+x_1x_2+x_2^2}{y_1^2+y_1y_2+y_2^2}=-\dfrac{x_1^2}{y_1^2}$$

    $$-\dfrac{x_1^2+x_1x_2+x_2^2}{y_1^2+y_1y_2+y_2^2}=-\dfrac{x_1^2}{y_1^2}$$

    $$x_1^2y_1^2+x_1x_2y_1^2+x_2^2y_1^2=x_1^2y_1^2+x_1^2y_1y_2+x_1^2y_2^2$$

    $$x_1x_2y_1^2+x_2^2y_1^2=x_1^2y_1y_2+x_1^2y_2^2$$

    $$x_1^2y_2^2-x_2^2y_1^2=x_1x_2y_1^2-x_1^2y_1y_2$$

    $$(x_1y_2-x_2y_1)(x_1y_2+x_2y_1)=x_1y_1(x_2y_1-x_1y_2)$$

    $$x_1y_2+x_2y_1=-x_1y_1$$

    $$\dfrac{x_2}{x_1}+\dfrac{y_2}{y_1}=-1$$
  • Question 2
    1 / -0
    The ordinate of all points on the curve $$y=\dfrac{1}{2\sin^{2}x+3\cos^{2}x}$$  where the tangent is horizontal, is
    Solution
    $$y=\dfrac{1}{2sin^{2}x+3cos^{2}x}=\dfrac{1}{2+cos^{2}x}$$
    $$tan\theta =\dfrac{dy}{dx}= \dfrac{-2sinx\, cos x}{(2+cos^{2}x)^{2}}=0$$
    $$\therefore sin\, x=0$$ or $$cos\, x= 0$$
    If $$sin\, x=0, y= \dfrac{1}{3}$$, if $$cos\, x=0, y= \dfrac{1}{2}$$

  • Question 3
    1 / -0
    The curve given by $$x + y = {e^{xy}}$$ has an tangents parallel to the y-axis at the point
    Solution
    $$\dfrac{dy}{dx}$$(differentiate the terms)
    $$1+\dfrac{dy}{dx}=e^{xy}\left(x\dfrac{dy}{dx}+y\right)$$
    $$\Rightarrow \dfrac{dy}{dx}=\dfrac{ye^{xy}-1}{1-xe^{xy}}$$
    Check by option
    $$\dfrac{dy}{dx}\left|_{(0, 1)}\right.=\dfrac{1-1}{1-0}=\dfrac{0}{1}=0$$ Not coming infinity
    $$\displaystyle\dfrac{dy}{dx}\left|_{(1, 0)}\right.=\dfrac{-1}{0}=\infty$$
    This point satisfy
    $$\therefore$$ option B$$(1, 0)$$.

  • Question 4
    1 / -0
    The slope of normal to the curve y= log (logx) at x = e is 
    Solution
    $$\begin{array}{l} \frac { d }{ { dx } } \log  \left( { \log  x } \right)  \\ Substitute\, u=\log  x \\ =\dfrac { d }{ { du } } \log  u\frac { d }{ { dx } } \log  x \\ =\dfrac { 1 }{ u } \left( { \frac { 1 }{ x }  } \right)  \\ =\dfrac { 1 }{ { \log  x } } \left( { \frac { 1 }{ x }  } \right)  \\ Substitute\, x=e \\ =\dfrac { 1 }{ { \log  e } } \left( { \dfrac { 1 }{ e }  } \right)  \\ =\dfrac { 1 }{ e }  \end{array}$$

  • Question 5
    1 / -0

    Number of possible tangents to the curve $$y = \cos \left( {x + y} \right), - 3\pi  \leqslant x \leqslant 3\pi $$, that are parallel to the line $$x + 2y = 0$$, is

    Solution

  • Question 6
    1 / -0
    The normal to the curve, $${x}^{2}+2xy-{3y}^{2}=0,\ at\left (1,1\right)$$:
    Solution

  • Question 7
    1 / -0
    Number of tangents drawn from the point $$\left (-1/2,0\right)$$ to the curve $$y={e}^{x}$$. (Here { } denotes fractional part function ). 
    Solution
    Let $$A$$ be the point of contact of a tangent of $${e}^{x}$$ passing through $$(-\cfrac{1}{2},0)$$
    $$\Rightarrow$$ $${ \left| \cfrac { dy }{ dx }  \right|  }_{ \lambda  }={ e }^{ \lambda  }$$
    Equation of tangent $$\quad y-{ e }^{ \lambda  }={ e }^{ \lambda  }(x-\lambda )\quad $$
    $$x=-\cfrac { 1 }{ 2 } ,y=0$$
    $$-{ e }^{ \lambda  }={ e }^{ \lambda  }\left( -\cfrac { 1 }{ 2 } -\lambda  \right) \Rightarrow { e }^{ \lambda  }\lambda =\cfrac { { e }^{ \lambda  } }{ 2 } $$
    either
    $$\quad { e }^{ \lambda  }=0$$ or $$\lambda =+\cfrac { 1 }{ 2 } $$
    $$\quad y{ e }^{ \lambda  }=0\Rightarrow y=0\quad \rightarrow$$ xaxis
    $$\quad \lambda =\cfrac { 1 }{ 2 } \Rightarrow y-{ e }^{ \cfrac { 1 }{ 2 }  }={ e }^{ \cfrac { 1 }{ 2 }  }\left( x-\cfrac { 1 }{ 2 }  \right) $$ 
    2 solutions or 2 distinct tangents
  • Question 8
    1 / -0
    If the tangent at P of the curve $$y^2=x^3$$ intersects the curve again at Q and the straight lines OP, OQ ma angles $$\alpha, \beta$$ with the x-axis where 'O' is the origin then $$\tan\alpha/\tan\beta$$ has the value equal to?
    Solution
    Let $$P$$ be $$(x_{1}, y_{1})$$ and $$Q$$ be $$(x_{2}, y_{2})$$
    equation of tangent $$(x_{2}, y_{2})$$, so $$\dfrac{y_{0}-y_{1}}{x_{2}-x_{1}}= \dfrac{3x_{1}^{2}}{2y_{1}}$$
    need to find out $$\dfrac{\tan \alpha}{ \tan \beta}  = \dfrac{\dfrac{y_{1}}{x_{1}}}{y_{2}/x_{2}} = \dfrac{y_{1}x_{2}}{x_{1} y_{2}}$$
    $$\dfrac{y_{2}-y_{1}}{x_{2}-x_{1}}= \dfrac{3x_{1}^{2}}{2y_{1}}$$
    $$2y_{1}y_{2}-2y^{2}= 3x^{2}_{1}x_{2}-3x_{1}^{3}$$
    $$2y_{1}y_{2}-2y_{1}^{2}= 3x_{1}^{2}x_{2}-3y^{2}$$
    $$2y_{1}y_{2}+y_{1}^{2}= 3x_{1}^{2} x_{2}$$
    $$y_{1}^{3} (2y_{2}+y_{1})^{3}= 27 x_{1}^{6} x^{3}_{2}$$
    $$y_{1}^{3} (2y_{2}+ y_{1})^{3}= 27 y_{1}^{4} y_{2}^{3}$$
    $$(2y_{2}+y_{1})^{3} - 27 y_{1} y_{2}$$
    $$8 y_{2}^{3} - 15 y_{2}^{2} y_{1}+ 6y_{2} y^{2}_{1}+ y_{1}^{3}=0$$
    $$(y_{2}-y_{1})^{2} (8y_{2}+ y_{1})=0$$
    $$8y_{2}=-y_{1} \Rightarrow 64 y_{2}^{2} = y_{1}^{2}$$
    $$64 x_{2}^{3} = x_{1}^{3} \Rightarrow 4 x_{2} = x_{1}$$
    $$\dfrac{\tan \alpha}{\tan \beta}= \dfrac{y_{1}x_{2}}{x_{1}y_{2}}$$
    $$= \dfrac{-8y_{2} x_{2}}{4x_{2} y_{2}} = -2$$
  • Question 9
    1 / -0
    A curve C has the property that if the tangent drawn at any point 'P' on C meets the coordinate axes at A and B, and P is midpoint of AB. If the curve passes through the point $$(1, 1)$$ then the equation of the curve is?
    Solution
    By option verification passing through $$(1,1)$$
    $$xy=1$$
  • Question 10
    1 / -0
    Let $$f$$ be continuous and differentiable function such that $$f(x)$$ and $$f^{'}(x)$$ have opposite sign everywhere. Then
    Solution

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